Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $a = \dfrac{-5x + 45}{x^2 - 4x - 45} \times \dfrac{6x + 30}{-3x - 12} $
Solution: First factor the quadratic. $a = \dfrac{-5x + 45}{(x + 5)(x - 9)} \times \dfrac{6x + 30}{-3x - 12} $ Then factor out any other terms. $a = \dfrac{-5(x - 9)}{(x + 5)(x - 9)} \times \dfrac{6(x + 5)}{-3(x + 4)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac{ -5(x - 9) \times 6(x + 5) } { (x + 5)(x - 9) \times -3(x + 4) } $ $a = \dfrac{ -30(x - 9)(x + 5)}{ -3(x + 5)(x - 9)(x + 4)} $ Notice that $(x - 9)$ and $(x + 5)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac{ -30(x - 9)\cancel{(x + 5)}}{ -3\cancel{(x + 5)}(x - 9)(x + 4)} $ We are dividing by $x + 5$ , so $x + 5 \neq 0$ Therefore, $x \neq -5$ $a = \dfrac{ -30\cancel{(x - 9)}\cancel{(x + 5)}}{ -3\cancel{(x + 5)}\cancel{(x - 9)}(x + 4)} $ We are dividing by $x - 9$ , so $x - 9 \neq 0$ Therefore, $x \neq 9$ $a = \dfrac{-30}{-3(x + 4)} $ $a = \dfrac{10}{x + 4} ; \space x \neq -5 ; \space x \neq 9 $